14 - Orthogonal

Orthogonal $n\times n$ matrix

$$A^T=A^{-1}$$

Properties of Orthogonal $n\times n$ Matrices

1. $AB$ and $A^{-1}$ are orthogonal
   Linking it to Term 2 Groups the $n\times n$ matrices form a group $O(n)$

2. $A$ is orthogonal $⟺$ the columns (or rows) are $n$ unit length, mutually perpendicular vectors

3. $A$ preserves the dot product i.e. If $\mathbf{x},\mathbf{y}\in {\mathbb{R}}_{col}^n$ then $A\mathbf{x}\cdot A\mathbf{y}=x\cdot y$

Proof

$(AB{)}^T=B^T{A}^T=B^{-1}{A}^{-1}=(AB{)}^{-1}$
$(A^{-1}{)}^T=(A^T{)}^T=A=(A^{-1}{)}^{-1}$
2

Preservation of Dot Product $⟺$ Orthogonal

$$A\text{ is orthogonal}⟺A\mathbf{x}\cdot A\mathbf{y}=x\cdot y\text{ for all }\mathbf{x},\mathbf{y}\in {\mathbb{R}}_{\mathrm{col}}^n$$

Proof

Using the fact that ${\mathbf{x}}^T\mathbf{y}=x\cdot y$

$$\begin{array}{rl}A\mathbf{x}\cdot \mathbf{A}\cdot \mathbf{y}& =\mathbf{x}\cdot \mathbf{y}\\ ⟺(A\mathbf{x}{)}^T(A\mathbf{y})& ={\mathbf{x}}^T\mathbf{y}\\ ⟺{\mathbf{x}}^T{A}^{T}A\mathbf{y}& =x^{T}y\end{array}$$

This is true if we can show that $A^{T}A=I_n$

By setting $\mathbf{x}={\mathbf{e}}_i$ and $\mathbf{y}={\mathbf{e}}_j$ (standard basis of ${\mathbb{R}}_{col}^n$)
Then $x^T{A}^{T}Ay$ selects the $(i,j)$th entry of $A^{T}A$
We also have $x^{T}y={\delta }_{ij}$ (Kronecker Delta)
Hence

$$(i,j)\text{th entry of }{A}^{T}A={\delta }_{ij}=(i,j)\text{th entry of }{I}_n$$

Therefore $A^{T}A=I_n$ hence ${\mathbf{x}}^T{A}^{T}A\mathbf{y}={\mathbf{x}}^T\mathbf{y}$